Zero Quantity Maximization

题目：

给定长度相等的数组a、b

c[i] = d*a[i] + b[i]

选择一个d使得c中0的个数尽可能多，问最多有多少0.

思路：

 考虑当a[i]为0时，只有b[i]也是0才满足条件

当a[i]不等于0时，计算b[i]/a[i]的值，用map记录并求解该键所对应的最大值

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include <bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<vector>
#include<cmath>
#include<string>
using namespace std;
typedef long long ll;
#define Maxn 200005
map<long double, ll>mp;
ll a[Maxn],b[Maxn];
ll n, res = 0, ans = 0;
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    //    c[i] = (-1) * b[i] / a[i];
    }
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) { //a[i]为0时，只有b[i]也是0,才能+1
            if (b[i] == 0) { //直接跳过
                res++;
            }
            continue;
        }
        mp[(long double)b[i] / a[i]]++;
        ans = max(ans, mp[(long double)b[i] / a[i]]);
    }
    
    cout << ans + res << endl;
    return 0;
}